Evaluate the following limits: x 9 sqrt(x)-3x-9 and t 0 sqrt(t2+9)-3t2.

$$\lim_{x \to 9} \frac{\sqrt{x}-3}{x-9}$$

Substituting $x=9$ directly gives $\frac{\sqrt{9}-3}{9-9} = \frac{3-3}{0} = \frac{0}{0}$, which is an indeterminate form.
To resolve this, we multiply the numerator and the denominator by the conjugate of the numerator, which is $(\sqrt{x}+3)$.

$$\lim_{x \to 9} \frac{\sqrt{x}-3}{x-9} \times \frac{\sqrt{x}+3}{\sqrt{x}+3}$$ $$\lim_{x \to 9} \frac{(\sqrt{x})^2 - 3^2}{(x-9)(\sqrt{x}+3)}$$ $$\lim_{x \to 9} \frac{x - 9}{(x-9)(\sqrt{x}+3)}$$

Since $x \to 9$, $x \neq 9$, so we can cancel the common factor $(x-9)$:

$$\lim_{x \to 9} \frac{1}{\sqrt{x}+3}$$

Now, substitute $x=9$:

$$\frac{1}{\sqrt{9}+3} = \frac{1}{3+3} = \frac{1}{6}$$

The value of the fourth limit is $\boxed{\frac{1}{6}}$.

$$\lim_{t \to 0} \frac{\sqrt{t^2+9}-3}{t^2}$$

Substituting $t=0$ directly gives $\frac{\sqrt{0^2+9}-3}{0^2} = \frac{\sqrt{9}-3}{0} = \frac{3-3}{0} = \frac{0}{0}$, which is an indeterminate form.
To resolve this, we multiply the numerator and the denominator by the conjugate of the numerator, which is $(\sqrt{t^2+9}+3)$.

$$\lim_{t \to 0} \frac{\sqrt{t^2+9}-3}{t^2} \times \frac{\sqrt{t^2+9}+3}{\sqrt{t^2+9}+3}$$ $$\lim_{t \to 0} \frac{(\sqrt{t^2+9})^2 - 3^2}{t^2(\sqrt{t^2+9}+3)}$$ $$\lim_{t \to 0} \frac{t^2+9 - 9}{t^2(\sqrt{t^2+9}+3)}$$ $$\lim_{t \to 0} \frac{t^2}{t^2(\sqrt{t^2+9}+3)}$$

Since $t \to 0$, $t \neq 0$, so we can cancel the common factor $t^2$:

$$\lim_{t \to 0} \frac{1}{\sqrt{t^2+9}+3}$$

Now, substitute $t=0$:

$$\frac{1}{\sqrt{0^2+9}+3} = \frac{1}{\sqrt{9}+3} = \frac{1}{3+3} = \frac{1}{6}$$